1. Rate Expression VIDEO AP 6.1

2. Collision Theory: When two chemicals react, their molecules have to collide with each other with proper energy and orientation. HOW FAST DOES A REACTION PROCEED?

3. Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

4. Reaction rate is the change in the concentration of a reactant or a product with time (M/s). This can be calculated using a rate expression in terms of reactants or products. A B rate = -  [A] tt rate =  [B] tt  [A] = change in concentration of A over time period  t  [B] = change in concentration of B over time period  t Because [A] decreases with time,  [A] is negative.

5. rate = -  [A] tt rate =  [B] tt

6. In this reaction, the concentration of chlorobutane, C 4 H 9 Cl, was measured at various times, t. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

7. The average rate of the reaction over each interval is the change in concentration divided by the change in time: C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq)

8.  Note that the average rate decreases as the reaction proceeds.  This is because as the reaction goes forward, there are fewer collisions between reactant molecules. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

9.  A plot of concentration vs. time for this reaction yields a curve like this.  The slope of a line tangent to the curve at any point is the instantaneous rate at that time. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

10.  In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1.  Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) Rate = -  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t

11. What if the ratio is not 1:1? H 2 (g) + I 2 (g)  2 HI (g) HI forms at half the rate H 2 disappears.

12. To generalize, for the reaction aA + bBcC + dD Reactants (decrease)Products (increase)

13. 1.Write the rate expressions: a.2H 2 + O 2  2H 2 O b.3O 2  2O 3 2.How is the rate of disappearance of O 2 related to the rate of appearance of O 3 if the O 3 rate equals 6.0x10 5 M/s? QUESTIONS 2.0x10 5 M/s -½ ΔH 2 /Δt = -ΔO 2 /Δt = ½ ΔH 2 O/Δt -3/2 ΔO 2 /Δt = ½ ΔO 3 /Δt -3/2 x = ½ (6.0x10 5 )

14. The student obtained the following data for the change of concentration of Br 2 : a. Calculate the average rate for the interval from 0 to 50s. b.Calculate the average rate for the interval from 50 to 100s. c.Explain why the rate is not the same for each time and how it was obtained. Time (s)[Br 2 ]Rate (M/s) 0.01204.20x10 -5 50.01013.52x10 -5 100.008462.96x10 -5 200.007102.49x10 -5 -3.80x10 -5 M/s -3.28x10 -5 M/s ΔBr 2 / Δ t (.0120-.0101)/(50)

15. Rate Laws VIDEO AP 6.2

16. The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. These powers are found experimentally and are referred to as orders. aA + bB cC + dD Rate = k [A] x [B] y

17. F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y Find a trial where F 2 changes and ClO 2 remains the same. F 2 doubles and the rate doubles, which is first order.

18. F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ] Find a trial where ClO 2 changes and F 2 remains the same. ClO 2 quadrupled and the rate quadrupled, which is 1 st order.

19.  If the concentration and the rate change the same way(1:1), it is first order.  If the concentration changes but the rate does not, it is zero order.  If the rate squares what the concentration does, it is second order.  The reaction order is the sum of all individual orders.  You may come across some questions with negative orders which means the rate changes the same magnitude the concentration does, just in the opposite direction. Rules

20. Run # Initial [A] ([A] 0 ) Initial [B] ([B] 0 ) Initial Rate (v 0 ) 11.00 M 1.25 x 10 -2 M/s 21.00 M2.00 M2.5 x 10 -2 M/s 32.00 M 2.5 x 10 -2 M/s What is the order with respect to A? What is the order with respect to B? What is the overall order of the reaction? 0 1 1

21. [NO (g) ] (mol dm -3 ) [Cl 2(g) ] (mol dm -3 ) Initial Rate (mol dm -3 s - 1 ) 0.250 1.43 x 10 -6 0.250 0.500 2.86 x 10 -6 0.500 1.14 x 10 -5 What is the order with respect to NO? What is the order with respect to Cl 2 ? What is the overall order of the reaction? 2 1 3

22. F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ] Summary of Rate Laws Rate laws are always determined experimentally. Reaction order is always defined in terms of reactant (not product) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. 1

23. [NO][H 2 ]Rate (M/s).0050.0020.0013.010.0020.0050.010.0040.010 Rate =k[NO] 2 [H 2 ] Overall order = 3.0013 = k[.005] 2 [.002] k = 26000/M 2 s FIND THE OVERALL RATE ORDER AND THE RATE CONSTANT:

24. [S 2 O 8 -2 ][I - ]Rate (M/s).080.034.00022.080.017.00011.160.017.00022 Rate =k[S 2 O 8 -2 ][I - ] Overall order = 2.00022 = k[.080][.034] k =.081/Ms A student is asked to run an additional trial using 2M S 2 0 8 -2 and 4M I -. Calculate the rate of this trial. FIND THE OVERALL RATE ORDER AND THE RATE CONSTANT:

25. Rate Equations VIDEO AP 6.3

26. First-Order Reactions: the rate depends on the concentration to the first power. rate = -  [A] tt rate = k [A] [A] t is the concentration of A at any time t [A] 0 is the initial concentration of A (at time t=0) ln[A] t - ln[A] 0 = - kt

27. The reaction 2A B is first order in A with a rate constant of 2.8 x 10 -2 s -1 at 80 0 C. How long will it take for A to decrease from 0.88 M to 0.14 M ln[A] - ln[A] 0 = - kt = 66 s [A] 0 = 0.88 M [A] = 0.14 M ln[0.14] – ln[.88] -2.8x10 -2

28. Cyclopropane changes to propene in a first order reaction with k=7.8x10 -4 s -1. If the initial concentration of cycloproane is 0.35M, calculate the final concentration after 9.8 min. ln[A] - ln[A] 0 = - kt ln[x] – ln [ 0.35 ] = -7.8x10 -4 (9.8*60) 0.22M Now, find out how long it takes to decrease the concentration from 0.75M to 0.05M. 3500s

29. First-Order Reactions The half-life, t ½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ? = 1200 s = 20. minutes How do you know decomposition is first order? units of k (s -1 ) ln[1] – ln[2] -5.7x10 -4

30. GRAPHICALLY: ln[A] - ln[A] 0 = - kt y - b = mx A plot of ln[A] vs t gives a straight line

31. FIRST-ORDER PROCESSES Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 NCCH 3 CN How do we know this is a first order rxn?

32. FIRST-ORDER PROCESSES When ln P is plotted against time, a straight line forms.  The process is first-order.  k is the negative slope: 5.1  10 -5 s -1.

33. Second-Order Reactions rate = -  [A] tt rate = k [A] 2 1 [A] - 1 [A] 0 = kt y - b = mx

34. A second order reaction with k = 7.0x10 9 /Ms at 23C. Find the final [I] after 22s if the initial equals 0.086M. Now, calculate the time it takes for [I] to change from 0.60M to 0.42M. 1/x -1/0.086 = (7.0x10 9 )(22) 6.5x10 -12 M 1/.42 -1/0.60 = (7.0x10 9 )(x) 1.0x10 -10 s 1 [A] - 1 [A] 0 = kt

35. DETERMINING REACTION ORDER Time (s)[NO 2 ], M 0.00.01000 50.00.00787 100.00.00649 200.00.00481 300.00.00380 NO 2 (g) NO (g) + 1/2 O 2 (g) Graphing ln [NO 2 ] vs. t yields: The plot is not a straight line, so the process is not first-order in [A].

36. A graph of 1/[NO 2 ] vs. t gives this plot. This is a straight line. Therefore, the process is second order in [NO 2 ]. K=.543/Ms

37. Zero-Order Reactions: not dependant on concentration. rate = -  [A] tt rate = k [A] 0 = k

38. Since rate is dependant on T, so is k.

39. Reaction mechanism VIDEO AP 6.4

40. The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. 2NO (g) + O 2 (g) 2NO 2 (g) N 2 O 2 is detected during the reaction! Elementary step:NO + NO N 2 O 2 Elementary step: N 2 O 2 + O 2 2NO 2 Overall reaction:2NO + O 2 2NO 2 +

41. Elementary step:NO + NO N 2 O 2 Elementary step:N 2 O 2 + O 2 2NO 2 Overall reaction:2NO + O 2 2NO 2 + Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step.

42.  Catalysts speed up the reaction and are found in the reactants of the first elementary step and the products of the last step. A + C  AC AC + B  ABC ABC  AB + C  A catalyst is homogeneous if it is in the same phase as the other reactants. If not, it is heterogeneous. Catalysts

44.  The RDS is always the slowest step.  It will have a similar Rate law as the overall reaction’s rate law. RATE DETERMINING STEP

45.  The rate law for this reaction is found experimentally to be Rate = k [NO 2 ] 2  CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration.  This suggests the reaction occurs in two steps. NO 2 (g) + CO (g)  NO (g) + CO 2 (g)

46.  A proposed mechanism for this reaction is Step 1: NO 2 + NO 2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast)  The NO 3 intermediate is consumed in the second step.  As CO is not involved in the slow, rate-determining step, it does not appear in the rate law. NO 2 (g) + CO (g)  NO (g) + CO 2 (g)

47. Write the rate law for this reaction.Rate = k [HBr] [O 2 ] List all intermediates in this reaction. List all catalysts in this reaction. HOOBr, HOBr None

48. CLASS SLIDES

49. Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- (aq) + 3I - (aq) 2SO 4 2- (aq) + I 3 - (aq) Experiment[S 2 O 8 2- ][I - ] Initial Rate (M/s) 10.080.0342.2 x 10 -4 20.080.0171.1 x 10 -4 30.160.0172.2 x 10 -4 rate = k [S 2 O 8 2- ] x [I - ] y X = 1 Y = 1 Overall = 2 rate = k [S 2 O 8 2- ] 1 [I - ] 1

50. [NO][Cl 2 ]Rate (M/min) 0.10 0.18 0.100.200.35 0.20 1.45 Rate =k[NO] 2 [Cl 2 ] Overall order = 3.18 = k[.1] 2 [.1] k = 180/M 2 min FIND THE OVERALL RATE ORDER AND THE RATE CONSTANT:

51. Time (s)[N 2 O 5 ] 00.91 3000.75 6000.64 12000.44 30000.16 FIND THE RATE CONSTANT GRAPHICALLY: Calculator steps: Stat Enter (edit) List t in L1 List Ln(N 2 O 5 ) in L2 Stat Right (calc) 4 Enter a = slope(m) = -k = -5.8x10 -4 k = 5.8x10 -4 /s

52. Time (s)Pressure (kPa) 0284 100220 150193 200170 250150 k= 2.6x10 -3 /s FIND THE RATE CONSTANT GRAPHICALLY:

53. Time (s)Pressure (kPa) 015.76 18112.64 5138.73 11644.44  First order r = -0.99957 m = -0.00108  Second Order r =.99138 m = 1.41x10 -4 First Order k =.00108/s DETERMINE THE ORDER AND RATE CONSTANT: